Linear Algebra and its Applications 332–334 (2001) 503–517www.elsevier.com/locate/laa
Index of Hadamard multiplication by positivematrices II
Gustavo Corach a ,∗,1, Demetrio Stojanoff b,2
aInstituto Argentino de Matemática, Saavedra 15 Piso 3 (1083), Buenos Aires, ArgentinabDepto. de Matemática, UNLP, 1 y 50 (1900), La Plata, Argentina
Received 23 October 1999; accepted 6 February 2001
Submitted by R.A. Horn
Abstract
For each n × n positive semidefinite matrix A we define the minimal index I (A)=max{λ �0 : A ◦ B � λB for all B � 0} and, for each norm N, the N-index IN (A) = min{N(A ◦ B) :B � 0 and N(B) = 1}, where A ◦ B = [aij bij ] is the Hadamard or Schur product of A =[aij ] and B = [bij ] and B � 0 means that B is a positive semidefinite matrix. A comparisonbetween these indexes is done, for different choices of the norm N. As an application we find,for each bounded invertible selfadjoint operator S on a Hilbert space, the best constant M(S)
Given A = [aij ], B = [bij ] ∈ Mn = Mn(C), the algebra of n × n matrices overC, denote by A ◦ B the Hadamard product [aij bij ]. In this paper A � 0 means
1 Supported by UBACYT TX79 and PIP 4463 (CONICET), Fundacion Antorchas and ANPCYT PICT97-2259 (Argentina).
2 Supported by UBACYT TW49 and PIP 4463 (CONICET) and ANPCYT PICT 97-2259 (Argentina).
0024-3795/01/$ - see front matter � 2001 Published by Elsevier Science Inc.PII: S 0 0 2 4 - 3 7 9 5 ( 0 1 ) 0 0 3 0 6 - 8
504 G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517
that A is positive semidefinite; Pn = {A ∈ Mn : A � 0} denotes the set of positivesemidefinite matrices.
Every A ∈ Mn defines a linear map �A : Mn → Mn given by �A(B) = A ◦ Bfor B ∈ Mn. By Schur’s product theorem [22] (see also [14, 7.5.3]) A ◦ B ∈ Pn ifA,B ∈ Pn so that �A is a positive linear map. Actually it is completely positive, i.e.,the inflation map �(m)
A , which acts entrywise as �A on Mm(Mn), is positive for allm ∈ N; see [20, Proposition 1.2].
In [23], the second author studied conditions under which
max{λ � 0 : �A(B) � λB, ∀B ∈ Pn
} = inf{‖�A(B)‖ : B ∈ Pn, ‖B‖ = 1
}.
The problem comes from the theory of conditional expectations. A conditional ex-pectation on a C∗-algebra A is a norm one projection E : A → A such that E(A)
is a sub-C∗-algebra of A. Every conditional expectation E satisfies the condition
sup{λ � 0 : ‖E(a)‖ � λ‖a‖ ∀a ∈ A+}= sup
{λ � 0 : E(a) � λa ∀a ∈ A+} , (1)
where A+ = {c ∈ A : c � 0}. The inverse of this number is called the index of Eand it is useful in the classification of inclusions of subalgebras of C∗-algebras. Notethat a conditional expectation is completely positive. If E : A → A is a completelypositive map that is not a conditional expectation, (1) fails in general and the problemarises of characterizing those E such that (1) holds.
For A ∈ Pn define the minimal index I (A) = max{λ � 0 : A ◦ B � λB ∀B ∈Pn} and the N-index IN(A) = max{λ � 0 : N(A ◦ B) � λN(B), ∀B ∈ Pn} for anygiven norm N on Mn. We are mainly concerned with Schatten norms ‖ · ‖p for p =1, 2, and ∞; we use the shorter notations I1, I2, and Isp for I‖·‖1 , I‖·‖2 , and I‖·‖∞ ,respectively. Isp is called the spectral index.
If A = Mn, every conditional expectation E has the form E(C) = U�A
(U∗CU)U∗, where U ∈ Mn is unitary and A ∈ Pn is a direct sum of matrices whoseentries are all equal to one. In this case, Ind(E)−1 = 1/k = Isp(A) = I (A), wherek is the number of diagonal blocks of A. We remove the inverse in our definition ofminimal and N-index in order to avoid complications when the index is zero.
For references on the norm of �A, see [2,3,9–11,17,19,20] and references in-cluded therein. There is an extensive bibliography about the index of conditionalexpectations; see [21] and its references. For a deep study of the index theory ofcompletely positive maps on operator algebras, see [5,12].
This paper compares these notions of index and investigates how to computethem. The results obtained are useful in the study of certain operator inequalities.Recall that if L(H) is the algebra of bounded linear operators on a Hilbert space Hand S ∈ L(H) is a selfadjoint invertible operator, then
‖ST S + S−1T S−1‖ � 2‖T ‖for all T ∈ L(H) [4]. It is natural to ask whether 2 is the best constant for each fixedS. Using a reduction to the finite dimensional case and a criterion for computing
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Isp(B) for matrices B ∈ Pn such that B � 0, in terms of the principal submatricesof B (see Corollary 4.6), we are able to find for each S, the best constant M(S) suchthat ‖ST S + S−1T S−1‖ � M(S)‖T ‖ for all T � 0.
In this paper we write A � 0 for matrices (or vectors) with nonnegative entries.We write A � B or A � B if A − B � 0 or A − B � 0, respectively. R(A) is therange of A and kerA is the kernel of A, where A is thought of as acting on Cn. AT
is the transpose matrix of A, A = [aij ] is the conjugate matrix of A, and A∗ = AT
.ρ(A) is the spectral radius of A and A† is the Moore–Penrose pseudoinverse of A.Throughout, p denotes the vector (1, . . . , 1)T and E denotes the matrix ppT, whichhas all its entries equal to 1.
Section 2 contains some elementary characterizations of the minimal index. Weprove that, for a given A ∈ Pn, I (A) > 0 if and only if p ∈ R(A); and, in this case,I (A)−1 is the spectral radius of A†E.
Section 3 is devoted to a comparison of the minimal index with the spectral index.The main result in this section is the following: if A ∈ Pn, A � 0, and there existsa vector u ∈ A−1({p}) such that u � 0, then I (A) = Isp(A). The converse holds ifI (A) /= 0, without the hypothesis that A � 0.
In Section 4 we compare the indexes associated with the spectral and the Frobeniusnorms. The main result here is that I2(A) = Isp(A ◦ A)1/2 for every A ∈ Pn. Asa consequence of the proof of this result we compute Isp(B) for matrices B ∈ Pnsuch that B � 0, in terms of the principal submatrices of B (see Corollary 4.6).This criterion is the main tool used in Section 5, where we compute the minimaland spectral indexes of ! = [λiλj + 1/λiλj ] for any n-tuple of positive numbersλ1, . . . , λn and use them to find, for each bounded Hermitian invertible operator Son a Hilbert space H, the number
M(S) = inf{‖ST S + S−1T S−1‖ : T � 0, ‖T ‖ = 1
}. (2)
For example, if ‖S‖ � 1, then M(S) = ‖S‖2 + ‖S‖−2.
2. Elementary properties of the index
Let us give more detailed definitions:
Definition 2.1. The Hadamard minimal index of A ∈ Pn is
I (A)=max{λ � 0 : A ◦ B � λB ∀B ∈ Pn
}=max
{λ � 0 : (�A − λ Id)B � 0 for all B ∈ Pn
}=max
{λ � 0 : A − λE � 0
}.
The last equality follows from the fact that for C ∈ Mn, the map �C is positive ifand only if C � 0.
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Definition 2.2. Given a norm N in Mn, the Hadamard N-index for A ∈ Pn is
IN(A)=max{λ � 0 : N(A ◦ B) � λN(B) ∀B ∈ Pn
}=min
{N(A ◦ B) : B ∈ Pn and N(B) = 1
}.
The index associated with the spectral norm ‖ · ‖ is denoted by Isp(·); we call it thespectral index. The index associated with the Frobenius norm ‖ · ‖2 is denoted byI2(·).
Example 2.3. Let A = [aij ] and B = [bij ] ∈ Pn. Then, if ‖ · ‖1 denotes the tracenorm,
‖B‖1 = tr(B) =n∑i=1
bii and ‖A ◦ B‖1 = tr(A ◦ B) =n∑
i=1
aiibii .
From these identities it is easy to see that, if I1(·) denotes the associated index, thenI1(A) = min1�i�n aii for every A ∈ Pn.
Remark 2.4. Estimation of the N-index of a matrix A can be seen as an inequality,namely, N(A ◦ B) � IN(A)N(B) for every B ∈ Pn. It would also be interesting toget such inequalities without the restriction B � 0 (of course, for matrices A withoutzero entries). But in this case, the constant involved is the inverse of the norm in-duced by N of the map �C , where cij = a−1
ij . The computation of such norms is wellknown (see [9–11,17,19,20]). For the index associated with the Frobenius norm, thecomputation of an infimum without the restriction B � 0 becomes trivial, but withthis restriction it is certainly not trivial, as shown in Theorem 4.3.
2.1. The minimal index I (A)
The index I (·) is called minimal because I (A) � IN(A) for every unitary invari-ant norm N. Indeed, given B ∈ Pn, then A ◦ B � I (A)B and, by Weyl’s monoton-icity theorem, si (A ◦ B) � I (A)si(B)), 1 � i � n (where si denote the ith singularvalue). ThereforeN(A ◦ B) � I (A)N(B) by Ky Fan’s dominance theorem; see [15,3.5.9].
Given B,C � 0 the following relation holds:
max{α � 0 : αC � B
} = ‖C1/2B†C1/2‖−1 = ρ(B†C)−1. (3)
In fact, if B is nonsingular, (3) follows from [14, 7.7.3] (see also [1,6,13,16]). If Bhas rank r < n, there exist a unitary matrix U and
� =[�1 00 0
]such that �1 is an r-by-r invertible matrix and B = U�U∗. If α � 0 and B � αC
G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517 507
D =[D11 D12D∗
12 D22
]= U∗CU,
we get
[�1 00 0
]= � � αD
so that D22 = 0 and, then, D12 = 0. Therefore �1 � αD11 and, by the nonsingularcase, ρ(�−1
1 D11) � α. The result follows by observing that ρ(�−11 D11) = ρ(B†C).
Observe also that the block structure of D and the invertibility of �1 imply theinclusion R(C) ⊂ R(B).
Taking B = A and C = E in (3) we get I (A) = max{α � 0 : A � αE} =ρ(A†E)−1 for every A ∈ Pn such that p ∈ R(A). This proves part of the followingresult.
Proposition 2.5. Let A ∈ Pn. Then I (A) /= 0 if and only if the vector p belongs toR(A). In this case, for any vector y such that Ay = p, we have
I (A) = ρ(A†E)−1 = 〈y, p〉−1 =(
n∑i=1
yi
)−1
. (4)
Proof. By definition, I (A) /= 0 if and only if there exists α > 0 such that A � αE.By the comments following (3), this means that R(E) ⊂ R(A) or, since p spansR(E), that p ∈ R(A). Finally, I (A)−1 = ρ(A†E) = ρ(A†ppT) = ρ(pTA†p) =pTA†p = 〈A†p,p〉, and A(A†p) = p. If y is any vector such that Ay = p, theny − A†p ∈ kerA = R(A)⊥, so 〈y, p〉 = 〈A†p,p〉. �
Proposition 2.6. Let A ∈ Pn. Then I (A) = min{〈z,Az〉 : ∑ni=1 zi = 1}.
Proof. If 〈z, p〉 = 1, then 〈z,Az〉 � I (A)〈z,Ez〉 = I (A) z∗pp∗z = I (A)〈z, p〉2 =I (A). If p ∈ R(A), let x ∈ Cn be such that Ax = p. Then z = I (A)x satisfies 〈z, p〉= I (A) 〈x, p〉 = 1 and 〈z,Az〉 = I (A)〈z, p〉 = I (A) by Proposition 2.5. Ifp /∈ R(A) = (kerA)⊥, then there exists z ∈ kerA such that 〈z, p〉 = 1 and 〈z,Az〉 =0 = I (A). �
Remark 2.7. Using Proposition 3.9 of [23] and the results of this section, it is easyto see that, for all A ∈ Pn and m ∈ N, the inflation matrix A(m) = Em ⊗ A (whereEm ∈ Pm has all its entries equal to 1) satisfies Isp(A
(m)) = Isp(A) and I (A(m)) =I (A). Note that the inflation map �(m)
A = �A(m) . Therefore the indexes of �A areinvariant under inflations and are invariants of �A as a completely positive map.
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2.2. IN(A) for general norms
Let A ∈ Pn and let N be a norm in Mn. If IN(A) = 0, there is some positivesemidefinite matrix C such that N(C) = 1 (so C /= 0) and N(A ◦ C) = 0. But thenA ◦ C = 0, so cij = 0 whenever aij /= 0. If all aii /= 0, then all cii = 0, which forcesC = 0. This contradiction shows that if all aii /= 0, then IN(A) > 0. Conversely, ifsome aii = 0 just take C = Eii/N(Eii), so I (A) � N(A ◦ C) = 0. Thus,
IN(A) > 0 if and only if all aii > 0. (5)
Let J ⊆ {1, 2, . . . , n} and let AJ denote the principal submatrix of A associated withJ. Then, minimality ensures that
IN(A) � IN (AJ ). (6)
Remark 2.8. Let A ∈ Pn. Then, it can be shown that for every unitary invariantnorm N, the following properties hold:1. If A has rank 1, then IN(A) = min1�i�n Aii .2. If A is positive and diagonal, then IN(D) = N ′(D−1)−1 = �′(a−1
11 , . . . , a−1nn )
−1,where N ′ is the dual norm of N and �′ is the symmetric gauge function on Rn
associated with N ′; see [7, Chapter IV].
Proposition 3.2 of [23] tells us that
Isp(A) = inf{Isp(D) : A � D and D is diagonal
}, (7)
and one could hope that a similar formula holds for any norm, but it does not. In fact,Corollary 4.4 says that, for every A ∈ Pn and the Frobenius norm,
I2(A)= inf
(
n∑1
D−2ii
)−1/2
: D is diagonal and A ◦ A � D2
= inf{I2(D) : D is diagonal and A ◦ A � D2
}. (8)
Note that the condition A ◦ A � D2 is strictly less restrictive than A � D (the re-verse implication follows from Schur’s theorem). Nevertheless, Eq. (8) allows oneto compute the 2-index for every positive semidefinite matrix using only diagonalmatrices. We intend to study this type of characterizations of IN(A) for generalnorms in a forthcoming paper.
3. I (A) = Isp(A)
In this section we characterize those matrices A ∈ Pn such that I (A) = Isp(A).In [23] it is shown that for
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A =(a b
b c
)∈ P2,
0 /= Isp(A) = I (A) ⇔ b ∈ R and 0 � b � min{a, c} /= 0. (9)
This is easily seen to be equivalent to the conditions1. A � 0.2. There exists a vector z � 0 such that Az = (1, 1)T (if A is invertible, this means
that A−1(1, 1)T � 0).We prove that, for positive semidefinite matrices of any size with nonnegative entries,condition 2 is equivalent to the identity Isp(A) = I (A). But first we need two lem-mas:
Lemma 3.1. Let A ∈ Pn and L = {z ∈ Rn : ∑i zi = 1}. Consider the sets
V1 = {z ∈ L : 〈Az, z〉 = I (A)
}and V2 = {
z ∈ L : Az = I (A)p}.
Then V1 = V2 /= ∅. Moreover, any local extreme point of the map G : L → R givenby G(z) = 〈Az, z〉, belongs to V2.
Proof. It is clear that V2 ⊆ V1. By Proposition 2.6, I (A) � min{〈Av, v〉: v ∈ L}.Then the mapG : L → R given by G(z) = 〈Az, z〉 = ∑
i,j aij zj zi is differentiableand bounded from below. Thus G must have a minimum, which is also a criticalpoint. Let the columns of X ∈ Mn,n−1 be a basis for the orthogonal complement ofp. Then we seek the unconstrained minimum of
over all ξ ∈ Rn−1. But ∇�(ξ) = 2XTA (Xξ + p/n) = 0 says that at a critical pointξ0, Az0 = �p for some �, where z0 ≡ Xξ0 + p/n ∈ L. But, in that case,
I (A) � 〈Az0, z0〉 = λ〈p, z0〉 = λ.
If I (A) = 0, then λ = 0, because p /∈ R(A), by Proposition 2.5. If I (A) > 0, thenalso λ = I (A), because y = λ−1z0 satisfies Ay = p and
λ = 〈Az0, z0〉 = λ2〈Ay, y〉 = λ2I (A)−1.
So ξ0 ∈ Rn−1 is a critical point of � if and only if z0 = Xξ0 + p/n ∈ V2. Sinceeach local extreme must be a critical point, this shows that ∅ /= V1 ⊆ V2 and thefinal assertion is proved. �
Lemma 3.2. Let A ∈ Mn, and suppose x ∈ Cn with ‖x‖ = 1. Let y = x ◦ x =(|x1|2, . . . , |xn|2)T.1. If Ay = �p for some λ ∈ C, then (A ◦ xx∗)x = λx.
510 G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517
2. Conversely, if all xi /= 0 and (A ◦ xx∗)x = λx for some λ ∈ C, then Ay = λp.If A ∈ Pn, the eigenvalue λ of the matrix A ◦ xx∗ associated with the vector x mustbe I (A) and Ay = I (A) p.
Proof. Suppose that Ay = �p. Then
(A ◦ xx∗)x=(aij xi xj )
x1...
xn
=(∑
j a1j |xj |2) x1...
(∑
j anj |xj |2) xn
=(Ay)1 x1
...
(Ay)n xn
= λx. (10)
Eq. (10) shows that if all xi /= 0 and (A ◦ xx∗)x = λx, then Ay = λp. If A ∈ Pn andI (A) = 0, then � = 0 because p /∈ R(A). If I (A) /= 0, then p ∈ R(A) = (kerA)⊥.SoAy /= 0 because 1 = ‖x‖2 = 〈p, y〉 /= 0. Then λ /= 0. If z = λ−1y, thenAz = p
and 1 = 〈p, y〉 = λ 〈Az, z〉 = λ I (A)−1, by Proposition 2.5, � = I (A). �
Theorem 3.3. Let A ∈ Pn.1. If Isp(A) = I (A) /= 0, then there exists a vector u � 0 such that Au = p.2. If A � 0 and there exists a vector u � 0 such that Au = p, then Isp(A) = I (A).
Proof.1. Observe that I (A)B � A ◦ B � ‖A ◦ B‖ I. By Lemma 2.1 of [23], there ex-
ists x ∈ Rn such that ‖x‖ = 1 and Isp(A) = ‖A ◦ xx∗‖. So, if y = x ◦ x, then〈y, p〉 = 1 and
I (A)xxT � DxADx � I (A)I,
which implies that
I (A) = I (A)(xTx)2 � xTDxADxx = yTAy � I (A)xTx = I (A).
We have 〈Ay, y〉 = I (A), y � 0 and 〈y, p〉 = 1. Then, by Lemma 3.1, Ay =I (A)p. Take u = I (A)−1y.
2. Let u be a nonnegative vector such that Au = p. Let y = I (A)u and x = (y1/21 ,
. . . , y1/2n )T. Note that ‖x2‖ = 〈y, p〉 = 1. By Lemma 3.2 we know that x is an
eigenvector of A ◦ xx∗ with eigenvalue I (A). Recall that always I (A) � Isp(A).
Case 1. Suppose that x has strictly positive entries. Since A ◦ xx∗ � 0, it is wellknown (see Corollary 8.1.30 of [14]) that the eigenvalue I (A) of x must be the
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spectral radius ofA ◦ xx∗. SinceA ◦ xx∗ ∈ Pn we deduce that I (A) = ‖A ◦ xx∗‖� Isp(A).
Case 2. Let J = {i : xi /= 0}, AJ the principal submatrix of A determined by theindexes of J and similarly define xJ . Then xJ is an eigenvector of AJ ◦ xJ x∗
J
with eigenvalue I (A). Note also that AJ ◦ xJ x∗J � I (AJ )xJ x
∗J and xJ x
∗J (xJ ) =
‖xJ ‖2xJ = xJ . Then
0 � 〈(AJ ◦ xJ x∗J − I (AJ )xJ x
∗J )xJ , xJ 〉 = I (A) − I (AJ )
and, by the definition of I, I (AJ ) = I (A). Now, as in Case 1, we can deduce that
I (A) = I (AJ ) = ‖AJ ◦ xJ x∗J‖ � Isp(AJ ) � Isp(A),
where the last inequality holds by (6). �
Corollary 3.4. Let A ∈ Pn such that A � 0 and Isp(A) = I (A). Let u be a nonneg-ative vector such that Au = p and y = I (A) u.1. Let x = (y
1/21 , . . . , y
1/2n )T. Then ‖x‖ = 1 and ‖A ◦ xx∗‖ = Isp(A).
2. Let J = {i : ui /= 0} and denote by AJ the principal submatrix of A determinedby J. Then I (A) = I (AJ ) = Isp(AJ ) = Isp(A).
Proof. This follows from the proof of Theorem 3.3. �
Remark 3.5. In Theorem 3.3(2), the hypothesis that A � 0 is essential. Indeed,consider
A =(
2 −1−1 1
)and u = (2, 3)T.
Then Au = (1, 1)T but 1/5 = I (A) /= Isp(A) = 1. For A ∈ Pn, we conjecture thatI (A) = I (sp,A) /= 0 implies that A � 0, as in the 2 × 2 case.
4. Isp(A) and I2(A)
In this section we study the relation between the indexes associated with thespectral and Frobenius norms. In Lemma 2.1 of [23] it is shown that the index Isp(·)is always attained at rank-1 projections. The index I1(·) has the same property (seeExample 2.3). It is natural to conjecture that the same result holds for any unitaryinvariant norm N. We show that the conjecture is true for the Frobenius norm:
Proposition 4.1. Let A ∈ Pn. Then there exists an x ∈ Cn such that ‖x‖ = 1 andI2(A) = ‖A ◦ xx∗‖2. That is, I2(A) is attained at a rank-1 projection.
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Proof. Let � = max{µ � 0 : ‖A ◦ B‖2 � µ‖B‖2 for all B ∈ Pn with rank 1}.By its definition � � I2(A). Let us prove that ‖A ◦ B‖2 � �‖B‖2 for all B ∈ Pn.Indeed, for B � 0, write B = ∑k
i=1 Bi , where each Bi has rank 1, Bi ∈ Pn, andBiBj = 0 if i /= j . Then
�2‖B‖22
= �2k∑
i=1
‖Bi‖22
�k∑
i=1
‖A ◦ Bi‖22.
On the other hand, using tr(XY ) � 0 for positive semidefinite matrices X and Y,
‖A ◦ B‖22 = tr((A ◦ B)∗(A ◦ B)) =
∑ij
tr(A ◦ Bi)∗(A ◦ Bj)
�∑i
tr(A ◦ Bi)∗(A ◦ Bi) =
∑i
‖A ◦ Bi‖22. �
Proposition 4.2. Let A ∈ Pn.1. There exists a nonnegative vector x such that ‖x‖ = 1 and ‖A ◦ xx∗‖2 = I2(A).2. Any such vector x satisfies (A ◦ A ◦ xx∗)x = I (BJ )x,whereB = A ◦ A and J =
{i : xi /= 0}.
Proof. Let y be a unit vector such that ‖A ◦ yy∗‖2 = I2(A). Let xi = |yi|. It is easilychecked that ‖x‖ = 1 and ‖A ◦ xx∗‖2 = ‖A ◦ yy∗‖2 = I2(A), which proves 4.2(1).Let B = A ◦ A ∈ Pn. Let y be a nonnegative unit vector and let z = (y2
1 , . . . , y2n)
T.Then
‖A ◦ yy∗‖22
=∑i,j
|aij |2y2i y
2j =
∑i,j
bij zj zi = 〈Bz, z〉
and∑n
1 zi = 1. Moreover, ‖A ◦ yy∗‖2 = I2(A) if and only if 〈Bz, z〉 is the min-imum of the map G(v) = 〈Bv, v〉 restricted to the simplex � = {v ∈ Rn : v � 0and
∑n1 vi = 1}. Using Lemma 3.1, we know that if z belongs to the interior �◦
of �, then z is a local extremum of G in the plane L = {z ∈ Rn : ∑i zi = 1 }, soBz = I (B) p.
If the vector x of item 1 satisfies xi > 0 for all i, then z = x ◦ x ∈ �◦ and Bz =I (B) p. By Lemma 3.2, (A ◦ A ◦ xx∗)x = I (B)x, showing 4.2(2) in this case. Ifsome xi = 0, let J = {i : xi /= 0}, let BJ be the principal submatrix of B determinedby the indexes of J, and similarly define xJ . Then I2(A) = ‖A ◦ xx∗‖2 = ‖AJ ◦xJ x
∗J‖2 � I2(AJ ) and
I2(A) = I2(AJ ) = ‖AJ ◦ xJ x∗J‖2,
because the converse inequality always holds by (6). Note that, by its definition,xJ has no zero entries. By the previous case, xJ is an eigenvector of BJ ◦ xJ x∗
J
with eigenvalue I (BJ ). But clearly B ◦ xx∗ has zeroes outside J × J , so x is aneigenvector of B ◦ xx∗ if and only if xJ is an eigenvector of BJ ◦ xJ x∗
J . Note thatthe eigenvalue of x is always I (BJ ). �
G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517 513
Theorem 4.3. Let A ∈ Pn. Then I2(A) = Isp(A ◦ A)1/2.
Therefore I2(A)2 � Isp(B). On the other hand, let x be a nonnegative unit vector
such that I2(A)2 = ‖A ◦ xx∗‖2 and J = {i : xi /= 0}. Then, by Proposition 4.2, (B ◦
xx∗)x = I (BJ )x and
I2(A)2 = ‖A ◦ xx∗‖2 = 〈(B ◦ xx∗)x, x〉 = I (BJ ).
But xJ is a unit eigenvector ofBJ ◦ xJ x∗J with strictly positive entries. So, by Lemma
3.2, BJ (xJ ◦ xJ ) = I (BJ )(1, . . . , 1)T. Suppose that I2(A) /= 0. Then I (BJ ) /= 0,BJ � 0, the vector u = I (BJ )
−1(xJ ◦ xJ ) has strictly positive entries, and BJu =(1, . . . , 1)T. Hence we can apply Theorem 3.3 to BJ and, by (6),
I (BJ ) = Isp(BJ ) � Isp(B) � I2(A)2 = I (BJ ).
If I2(A) = 0, then (5) ensures that some aii = 0, so Isp(B) = 0 by (5). �
Corollary 4.4. Let A ∈ Pn. Then
I2(A)= inf
(
n∑1
d−2ii
)−1/2
: D is positive diagonal and A ◦ A � D2
= inf{I2(D) : D is positive diagonal and A ◦ A � D2
}.
Proof. This is a direct consequence of Theorem 4.3 and Proposition 3.2 of [23]. �
Remark 4.5. In Theorem 4.3 we get information about A ∈ Pn using B = A ◦ A.But it can also be used to get information about any B ∈ Pn with B � 0, using A =(b
1/2ij ). Unfortunately it may certainly happen thatA /∈ Pn. Nevertheless this obstruc-
tion can be removed in the following way: given a selfadjoint (but not necessarilypositive semidefinite) matrix A ∈ Mn, consider the index
I2(A) = min{‖A ◦ xx∗‖2 : |x‖ = 1
},
which, by Proposition 4.1, is consistent with Definition 2.2 when A � 0. A carefulinspection of the proofs of Proposition 4.2 and Theorem 4.3 shows that they remaintrue using this new index if the condition “A ∈ Pn” is replaced by “A = A∗ and B =A ◦ A ∈ Pn”. Note that Lemmas 3.1 and 3.2, and Theorem 3.3 are applied only to thepositive semidefinite matrix B or its principal submatrices. The inequality I2(A) �I2(AJ ) in (6) (which is also used in the proofs) remains valid for this new index. Thisobservation is useful in order to avoid the unpleasant condition “A = (b
514 G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517
Corollary 4.6. Suppose B ∈ Pn and B � 0. Then there exists a subset J0 of{1, 2, . . . , n} such that Isp(B) = Isp(BJ0) = I (BJ0). Therefore
Isp(B) = min{Isp(BJ ) : Isp(BJ ) = I (BJ )
}.
If A = (b1/2ij ) (which may be not positive semidefinite), then J0 can be characteri-
zed as J0 = {i : xi /= 0} for any unit vector x such that I2(A) = ‖A ◦ xx∗‖2. AlsoIsp(B) = ‖B ◦ xx∗‖ = 〈By, y〉, where y = (|x1|2, . . . , |xn|2)T .
Proof. Use Remark 4.5 and the proof of Theorem 4.3. �
5. An operator inequality
In this section we compute the indexes of a particular class of matrices and, asan appplication, we get a new operator inequality, closely related to the inequalityproved in [8]; see also [4,18].
Let x = (λ1, . . . , λn)T ∈ Rn+, S = {λ1, . . . , λn}, and
� = �x =(λiλj + 1
λiλj
)ij
∈ Pn.
Observe that � has rank 1 or 2.
5.1. Computation of I (�)
1. If all λi are equal, then � = (λ21 + λ−2
1 ) E and I (�) = λ21 + λ−2
1 .2. If #S > 1, then the range of � is spanned by the independent vectors x =
G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517 515
5.2. Computation of Isp(�)
We shall compute Isp(�) using Corollary 4.6 and therefore use the principal sub-matrices of �, which are matrices of the same type. Let J ⊂ {1, 2, . . . , n}, let SJ ={λj : j ∈ J }, and let xJ be the induced vector. Then �J = �xJ and so Isp(�J ) /= 0.Suppose that Isp(�J ) = I (�J ). Then #SJ � 2 by Section 5.1. If #SJ = 2, let i1, i2 ∈J be such that λi1 /= λi2 . By Theorem 3.3 there exists a vector y ∈ RJ such thaty � 0 and �J y = pJ . Let z1 = ∑{yk : k ∈ J and λk = λi1 } � 0 and z2 = ∑{yj :j ∈ J and λj = λi2 } � 0. Easy computations show that �{i1,i2}(z1, z2)
T = (1, 1)T.Then, by Theorem 3.3 and Section 5.1,
Isp(�J ) = I (�J ) = (λi1 + λi2)2
1 + λ2i1λ2i2
= I (�{i1,i2}) = Isp(�{i1,i2}).
Therefore, in order to compute Isp(�) using Corollary 4.6, we need to consider onlythe diagonal entries of � and some of the principal submatrices of size 2 × 2. Ifλi /= λj , (9) ensures that
Isp(�{i,j}) = I (�{i,j}) ⇔ λiλj + 1
λiλj� min
{λ2i + 1
λ2i
, λ2j + 1
λ2j
}.
If λi < λj , this condition is equivalent to
λ2i � 1
λiλj� λ2
j . (11)
In particular, this implies that λi < 1 < λj . Then, by Corollary 4.6,
Isp(�) = min{M1,M2} (12)
where M1 = mini λ2i + λ−2
i = mini �ii and
M2 = inf
{(λi + λj )
2
1 + λ2i λ
2j
: λi < 1 < λj and λ2i � 1
λiλj� λ2
j
}.
For example, if all λi � 1 (or all λi � 1), then by (11), Isp(�) = M1 = mini λ2i +
λ−2i . On the other hand, if λ /= 1 and x = (λ, λ−1)T, then
Isp(�x) = M2 = λ2 + λ−2
2+ 1 < M1 = λ2 + λ−2.
Proposition 5.1. Let H be a Hilbert space and let S be a bounded selfadjointinvertible operator on H. Let M(S) be the best constant such that
‖ST S + S−1T S−1‖ � M(S)‖T ‖ for all 0 � T ∈ L(H).
Then M(S) = min{M1(S),M2(S)}, where
M1(S) = minλ∈ σ(S)
λ2 + λ−2
516 G. Corach, D. Stojanoff / Linear Algebra and its Applications 332–334 (2001) 503–517
Proof. We follow the same steps as in [8]. By taking the polar decomposition of S,we can assume that S > 0, because the unitary part of S is also the unitary part ofS−1; it commutes with S and S−1 and it preserves norms. Note that we must changeσ(S) by σ(|S|) = {|λ| : λ ∈ σ(S)}.
By the spectral theorem, we can assume that σ(S) is finite, because S can beapproximated in norm by operators Sn such that each σ(Sn) is a finite subset ofσ(S), σ(Sn) ⊂ σ(Sn+1) for all n ∈ N and
⋃n σ (Sn) is dense in σ(S). Then M(Sn)
(and Mi(Sn), i = 1, 2) converges to M(S) (resp. Mi(S), i = 1, 2).We can suppose also that dimH < ∞, by choosing an appropriate net of finite
rank projections {PF }F∈F that converges strongly to the identity and replacing S, Tby PFSPF , PF T PF . Indeed, the net may be chosen in such a way that SPF = PF S
and σ(PF SPF ) = σ(S) for all F ∈ F. Note that for every A ∈ L(H), ‖PFAPF ‖converges to ‖A‖.
Finally, we can suppose that S is diagonal by a unitary change of basis in Cn. Inthis case, if λ1, . . . , λn are the eigenvalues of S (with multiplicity) and x =(λ1, . . . , λn)
T, then
ST S + S−1T S−1 = �x ◦ T .None of our reductions (unitary equivalences and compressions) change the fact that0 � T . Now the statement follows from formula (12). If ‖S‖ � 1, then M(S) =M1(S), because M2(S) is the infimum of the empty set. Clearly M1(S) is attained atthe element λ ∈ σ(S) such that |λ| = ‖S‖. �
Acknowledgement
We thank the referees, who helped us to improve several statements and proofsthroughout the paper. In particular, the present proofs of Lemma 3.1, item 1 ofTheorem 3.3, and the inclusion of the spectral radius in Proposition 2.5 are due tothem.
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